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JEE PHYSICS MCQ: DUAL NATURE OF RADIATION AND MATTER

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The work function for caesium is 1.8 eV1.8 \mathrm{~eV}. Light of 5000 A∘5000 ~\mathop {\text{A}}\limits^ \circ is incident on it. Calculate maximum kinetic energy of the emitted electrons. Given : h=6.6×10−34 Jsh=6.6 \times 10^{-34} \mathrm{~Js}, me=9×10−31 kg,c=3×108 ms−1m_e=9 \times 10^{-31} \mathrm{~kg}, c=3 \times 10^8 \mathrm{~ms}^{-1}.
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Correct option: A
1.08×10−19 J1.08 \times 10^{-19} \mathrm{~J}
Option A (1.08×10−19 J1.08 \times 10^{-19} \mathrm{~J}) is correct because it is the maximum kinetic energy calculated using the photoelectric effect equation.

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